SOLUTION: Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well. A rectangular ga

Click here to see ALL problems on Quadratic Equations

Question 184471: Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well.
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 15ft by 25ft. The total area, garden plus walkway, is to be 650ft^2. What must be the width of the walkway?

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A rectangular garden is to be surrounded by a walkway of constant width.
The garden's dimensions are 15ft by 25ft.
The total area, garden plus walkway, is to be 650ft^2.
What must be the width of the walkway?
:
Let x = width of the walkway
:
Draw a diagram of this and it will be apparent that the total area dimensions are:
(2x+15) by (2x+25) and that equals 650 sq/ft
:
A simple area equation
(2x+15)*(2x+25) = 650
FOIL and subtract 650 from both sides:
4x^2 + 50x + 30x + 375 - 650 = 0
:
4x^2 + 80x - 275 = 0; our old friend, the quadratic equation!
:
Use the quadratic formula to find x

In this equation, a=4, b=80, c=-275

:

:

Two solutions, but we are ignoring the negative solution

x =
x = 2.99 ft wide is the walkway
:
:
Check solution in the area equation, add 2 * 2.99 = 5.98 to each dimension
30.98 + 20.98 = 649.96 ~ 650